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SSC CGL Mains Geometry Questions : 1st July

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As you all know, the upcoming months are lined up with various important exams like SSC CGL Mains 2018, so we are here to help you with the subject that is common to all of the given exams. We are providing daily quantitative aptitude quizzes, practice which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.

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Q1. In the given figure, a circle touches the sides of the quadrilateral PQRS. The radius of the circle is 9 cm. ∠RSP = ∠SRQ = 60° and ∠PQR = ∠QPS = 120°. What is the perimeter (in cm) of the quadrilateral ?
दी गयी आकृति में, एक वृत्त चतुर्भुज PQRS के भुजाओं को छूता है। वृत्त की त्रिज्या 9 सेमी है। ∠RSP = ∠SRQ = 60° और ∠PQR = ∠QPS = 120°. चतुर्भुज की परिधि (सेमी में) क्या है?

(a) 36√3
(b) 24√3
(c) 48√3
(d) 32
Ans.(C)
Sol.
Here given a diagram as shown in the question

∠RSP = ∠SRO = 60 (Given)
∠PQR = ∠QPS = 120 (Given)
In a quadrilateral PMOK
∠MOK = 180 – 120 = 60
∠POM = ∠KOP = 30
OK = OM = 9 (Given)
In ∆POM
tan⁡30=PM/9 ⇒PM=3√3=PK=MQ=QL

Similarly In ∆SON
tan⁡30=ON/SN
SN = 9√3 = SK = NR = RL
So, perimeter PQRS
= 3√3×4+9√3×4
= 4 × 12√3
= 48√3

Q2.If the following figure, find the value of x
निम्नलिखित आकृति में, x का मान ज्ञात कीजिए

(a) 40°
(b) 45°
(c) 50°
(d) 60°
Ans.(c)
Sol.
∠BDC = 30°
∠BAC = ∠BDC (∵ both angle are on same arc)
∠BAC = 30°
∠ACB = x° = 180° – (100° + 30°)
x° = 50°

Q3. In the given figure, from the point P two tangents PA and PB are drawn to a circle with centre O and radius 5 cm. From the point O, OC and OD are drawn parallel to PA and PB respectively. If the length of the chord AB is 5 cm, then what is the value (in degrees) of ∠COD?
दिए गए आंकड़े में, बिंदु P से दो स्पर्शरेखाएँ PA और PB केंद्र O और त्रिज्या 5 सेमी के एक वृत्त पर खींची गई हैं। बिंदु O से, OC और OD क्रमशः PA और PB के समानांतर खींचे जाते हैं। यदि जीवा AB की लंबाई 5 सेमी है, तो ∠COD का मान (डिग्री में) क्या है?

(a) 90
(b) 120
(c) 150
(d) 135
Ans.(B)
Sol.

AB = 5 (Given)
Radius = 5 (given)
Hence ∆AOB is an equilateral triangle,
So,
θ = 60° = ∠BOM
because AO = OB = radius
∠AOB = 60
∠PAM = ∠PBM = 90 – 60 = 30 each
∠APB = 180 – (30 + 30) = 120 = ∠COD
Because PA∥OC & PB∥OD

Q4. In the given figure, Tangents TQ and TP are drawn to the larger circle (centre O) and tangents TP and TR are drawn to the smaller circle (centre O’). Find TQ : TR :–
दिए गए आंकड़े में, स्पर्शरेखा TQ और TP को बड़े वृत्त (केंद्र O) पर खींचा जाता है और स्पर्श रेखाओं TP और TR को छोटे वृत्त (केंद्र O ‘) पर खींचा जाता है। TQ: TR ज्ञात करें :-

(a) 1 : 1
(b) 5 : 4
(c) 8 : 7
(d) 7 : 8
Ans.(a)
Sol.
TQ = TP (∵ Both are tangents on circle from same point)
And also
TR = TP
∴ TQ = TR
⇒ TQ : TR = 1 : 1

Q5. In the given figure, AB is a diameter of the circle with centre O and XY is the tangent at a point C. If ∠ACX = 35°, then what is the value (in degrees) of ∠CAB ?
दिए गए आंकड़े में, AB, केंद्र O के वृत्त का व्यास है और XY एक बिंदु C पर स्पर्शरेखा है। यदि ∠ACX = 35° है, तो ∠CAB का मान (डिग्री में) क्या है?

(a) 45
(b) 35
(c) 55
(d) 65
Ans.(C)
Sol.

∠ACB = 90 [angle on semicircle is always 90]
∠ACX = 35
∠ACX = ∠ABC = 35° (it is a property)
So,
∠BAC or ∠CAB = (90 – 35) = 55°

Q6. P and Q are the middle points of two chords (not diameters) AB and AC respectively of a circle with centre at a point O. the lines OP and OQ are produced to meet the circle respectively at the points R and S. T is any point on the major arc between the points R and S of the circle. If ∠BAC = 32°, ∠RTS = ?
P और Q दो जीवाओं AB और AC का मध्य बिंदु हैं (व्यास नहीं) बिंदु O वृत्त का केंद्र हैं। OP और OQ लाइनों को R और S के बिंदुओं पर क्रमशः वृत्त से मिलने के लिए बढाया जाता है। T, वृत्त पर R और S के बीच प्रमुख चाप पर कोई बिंदु है। यदि ∠BAC = 32°, ∠RTS = ?
(a) 32°
(b) 74°
(c) 106°
(d) 64°
Ans.(b)
Sol.

P and Q are mid points of AB and AC so OP and OS will be perpendicular at AB and AC respectively.
∴ ∠APO = ∠AQO = 90°
⇒ ∠ROS = 180° – 32°
∠ROS = 148°
∠RTS = 1/2 ∠ROS
∠RTS = 74°

Q7. In the given figure, PQ is a diameter of the semicircle PABQ and O is its center. ∠AOB = 64°. BP cuts AQ at X. What is the value (in degrees) of ∠AXP ?
दिए गए आंकड़े में, PQ अर्धवृत्त PABQ का एक व्यास है और O इसका केंद्र है। ∠AOB = 64° है. BP, AQ को X पर काटती है। ∠AXP का मान (डिग्री में) क्या है?

(a) 36
(b) 32
(c) 58
(d) 54
Ans.(c)
Sol.GIVEN DIAGRAM PQ is diameter of circle with center O, angle AOB=64

If we draw line AP Then
∠APB = 1/2∠AOB =32 (Angle at circumference with same arc is always half of the angle at center)
∠PAQ = 90 (angle at semicircle with diameter is always 90)
So,
∠AXP = 180 – (∠APB+∠PAQ)=180-(90+22) = 58°

Q8.In the given figure, find the radius of smaller circle (r) :
दी गयी आकृति में, छोटे वृत्त की त्रिज्या ज्ञात करें

(a) (√3-2√2)R
(b) 2(√2-1)R
(c) (3-2√2)R
(d) None of these
Ans.(c)
Sol.

Radius of larger circle = R
Radius of smaller circle = r
Centre of smaller circle = c
OA =R√2
AC =r√2
OA = R + r + r√2
R√2=R+r(√2+1)
R(√2-1)=r(√2+1)
r=((√2-1)/(√2+1))R or (3-2√2)R

Q9. In the given figure, E and F are the centers of two identical circles. What is the ratio of area of triangle AOB to the area of triangle DOC ?
दिए गए आंकड़े में, E और F दो समान वृत्त के केंद्र हैं। त्रिभुज AOB और त्रिभुज DOC के क्षेत्रफल का अनुपात क्या है?

(a)1 : 3
(b) 1 : 9
(c)1 : 8
(d) 1 : 4

Ans.(B)
Sol. ACCORDING TO QUESTION FIGURE IS DRAWN, two circles with center E & F

∠AEB ≈ MED
∠AFB ≈ CFN
So,
AB = DM = CN
&
AB = MN
[As shown in fig]
AB/(DM+MN+NC)=AB/DC=1/3

So,
(Area of ∆AOB)/(Area of ∆DOC)=(AB^2)/(DC^2 )=(1/3)^2=1/9

Q10.In the given figure, ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A, B and C. If ∠ADC = 130°. Find ∠CAB.
दिए गए आंकड़े में, ABCD एक चक्रीय चतुर्भुज है जिसका भुजा AB वृत्त पर बिंदु A, B और C के माध्यम से वृत्त का व्यास है यदि ∠ADC = 130 °. ∠CAB का मान ज्ञात करें

(a) 40°
(b) 50°
(c) 30°
(d) 130°
Ans.(b)
Sol.
∠ADC=130°
∴∠ABC=180°-∠ADC
∠ABC=50°
∠CAB=90°-∠ABC
∠CAB=90°-50°=40°

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